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I'm a high-school student of physics. While solving some problems from here, I came across this:

Consider three charges $q_1$, $q_2$ and $q_3$ each equal to $q$ at the vertices of an equilateral triangle of side $l$. What is the force on a charge $Q$ (with the same sign as $q$) placed at the centroid of the triangle, as shown in Fig. 1.9?

I know how to solve this problem using Coulomb's law and a little vector analysis, and that's what I did. The answer's $\vec{0}$. However, at the end of the solution (the problem's been given as an example), this was written:

It is clear also by symmetry that the three forces will sum to zero. Suppose that the resultant force was non-zero but in some direction. Consider what would happen if the system was rotated 60° about $O$

I don't quite understand what role symmetry has to play here. As a matter of fact, I'm awfully puzzled by the very notion of symmetry in physics. I'd like to know how we can answer this question using just symmetry, and how do I understand symmetry in the context of physics, so that I can identify and solve correctly more such problems using symmetry with little to no algebra.

PS: (1) Here's Fig 1.9: picture of an equilaterial triangle with vertices A, B, and C, where each vertex has a charge Q. The centroid is labelled O and a charge Q sits on it. A perpendicular AD from A to BC is drawn, and the side lengths are labelled 'l' (2) The concerned problem is on page 16 of the PDF linked above.

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    $\begingroup$ An easier example: imagine a uniformly charged spherical shell. What would be the electric field at the center? The sphere is fully symmetric, it would make no sense that the electric field would point at a specific direction, as the charge configuration looks the same in any direction (it is invariant under any rotation). Which one would you chose? . So it must be zero at the center. $\endgroup$ Oct 19, 2023 at 3:47

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The basic principle of symmetry being used here is that if the problem has a certain symmetry, then the solution must also have that symmetry.

In this case, if you rotate the charges by 120 degrees about the center point of the triangle, then you end up with the same configuration of charges. Therefore, the problem has a symmetry under rotations by 120 degrees about the center point of the triangle.

Now, suppose the solution involved the electric field pointing in, say, the "up" direction (toward the charge at the top of the triangle). Now again imagine rotating the configuration of charges by 120 degrees. The electric field vector will now be pointing diagonally downward. But this is a contradiction -- for a given configuration of charges, there is only one possible electric field. Yet, we have shown that there are two allowed electric fields for the same problem. This is a contradiction. The resolution is that the electric field must have been zero.

In other words, the original problem does not change if we rotate the charges by 120 degrees. Therefore, the solution (ie, the electric field) must also not change if we rotate the charges by 120 degrees. In this context, the only possible electric field is zero, since any non-zero vector will change under a 120 degree rotation.

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  • $\begingroup$ @PatoGalmarini Gah. Thanks for the correction. $\endgroup$
    – Andrew
    Oct 19, 2023 at 3:57
  • $\begingroup$ Thanks for the answer @Andrew. I'd also like to clarify something further—let's assume that there's no charge this time at the centroid, and the A vertex has a negative charge of the same magnitude, and the charges at B and C remain the same. Now, the forces on each of the charges at A, B and C would obviously change, but there'd be bilateral symmetry with AD as the axis. So, by this logic, can we say that the forces on B would mirror that of C and vice versa? $\endgroup$ Oct 19, 2023 at 5:38
  • $\begingroup$ "The basic principle of symmetry being used here is that if the problem has a certain symmetry, then the solution must also have that symmetry." I think this should be rephrased, I don't how but should be, because this young man will soon encounter Bloch waves and will be as puzzled by its non-periodicity as I was because I also thought that geometrical symmetry must mean an equal amount of solution symmetry. $\endgroup$
    – hyportnex
    Oct 19, 2023 at 11:42
  • $\begingroup$ @hyportnex That's fair. The phrase "being used here" was meant to imply this isn't the full story, since as you say symmetries in quantum mechanics are more subtle than in classical electromagnetism. There probably is a better way to phrase this though. I'll think about it. $\endgroup$
    – Andrew
    Oct 19, 2023 at 14:19
  • $\begingroup$ @Andrew "The basic principle of symmetry being used here is that if the problem has a certain symmetry, then the solution must also have that symmetry." Where are these principles discussed? $\endgroup$
    – user45664
    Oct 19, 2023 at 18:17

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